SINGULAR HARDY-TRUDINGER-MOSER INEQUALITY AND THE EXISTENCE OF EXTREMALS ON THE UNIT DISC

. We present the singular Hardy-Trudinger-Moser inequality and the existence of their extremal functions on the unit disc B in R 2 . As our ﬁrst main result, we show that for any 0 < t < 2 and u ∈ C ∞ 0 ( B ) satisfying there exists a constant C 0 > 0 such that the following inequality Furthermore, by the method of blow-up analysis, we establish the existence of extremal functions in a suitable function space. Our results extend those in Wang and Ye [36] from the non-singular case t = 0 to the singular case for 0 < t < 2.

(Communicated by Guozhen Lu) Abstract. We present the singular Hardy-Trudinger-Moser inequality and the existence of their extremal functions on the unit disc B in R 2 . As our first main result, we show that for any 0 < t < 2 and u ∈ C ∞ 0 (B) satisfying there exists a constant C 0 > 0 such that the following inequality holds Furthermore, by the method of blow-up analysis, we establish the existence of extremal functions in a suitable function space. Our results extend those in Wang and Ye [36] from the non-singular case t = 0 to the singular case for 0 < t < 2.
Theorem A. Let u ∈ W 1,n 0 (Ω), for 0 ≤ α ≤ α n = nω 1 n−1 n−1 , where ω n−1 denotes the surface area of the unit sphere in R n , then there exists a positive constant C n such that the following inequality holds sup ∇u L n (Ω) ≤1 Ω exp(α |u| n n−1 )dx ≤ C n |Ω|. (1.1) The constant α n is sharp in the sense that if we replace α > α n , then the inequality (1.1) can no longer hold with some C n independent of u.
There has been a lot of research about the Trudinger-Moser inequalities. Some results have played fundamental roles in solving partial differential equations. Here, we only recall some related results.
In 2012, Wang and Ye [36] established the following Hardy-Trudinger-Moser inequality on the unit disc B in R 2 by using the Schwartz rearrangement argument. They also proved the supremum can be achieved in a suitable function space by blow-up analysis. (1−|z| 2 ) 2 dxdy, then there exists a constant C > 0 such that Furthermore, the supremum can be achieved in a suitable function space.
We can observe the second term of the H(u) is the square of L 2 norm on the hyperbolic disk. Thus, it is closely related to the Trudinger-Moser inequality on the hyperbolic spaces (see Mancini and Sandeep [31], and Lu and Tang [23,24], etc.).
Wang and Ye conjectured the Hardy-Trudinger-Moser inequality will be true for any bounded and convex domain with smooth boundary. Lu and Yang gave an affirmative answer to this conjecture. They confirmed this conjecture indeed holds for any bounded and convex domain in R 2 in [25]. They proved Theorem C via an argument from local inequalities to global ones using the level sets of functions under consideration developed by Lam and Lu in [13,14] (see also [5,16,39]), together with the Riemann mapping theorem.
Theorem C. Let Ω be a bounded and convex domain in R 2 . Then there exists a constant C(Ω) > 0 such that for any u ∈ C ∞ 0 (Ω), the following inequality holds Later, Mancini, Sandeep and Tintarev proved another modified Trudinger-Moser inequality on B in [32] by rearrangement. They proved for all u ∈ C ∞ 0 (B) satisfying then there exists a constant C > 0 such that the following inequality holds Theorems B and Theorems C of the Hardy-Trudinger-Moser type inequalities have been established recently in higher dimensional cases for higher order derivatives on hyperbolic balls B n using Fourier analysis on hyperbolic spaces and Greens' function estimates, namely, the Hardy-Adams inequalities which improve the classical Adams inequalities on balls [1] when the dimension n ≥ 4 and is even ( [15,26]). (see also for non-compact Riemanian manifolds with negative curvatures [8,37]).
These Hardy-Adams inequalities are the borderline case of the high order Hardy-Sobolev-Maz'Ya inequality proved in [27].
Employing a rearrangement argument and a change of variables, Adimurthi and Sandeep generalized the Trudinger-Moser inequality (1.1) to a singular version in [2]. They proved the following theorem.
Theorem D. Let Ω ⊂ R n , α ≥ 0, 0 ≤ β < n satisfy α αn + β n ≤ 1, then the following inequality holds This inequality was extended to Ω = R n in Adimurthi and Yang [3] which gave a singular version of the inequality in Li and Ruf [21]. Then, Dong and Lu [7], and Dong, Lam and Lu [6] established the two weighted Trudinger-Moser inequality in R n . (See also Lam and Lu [11,12] for more singular Trudinger-Moser inequalities on R n .) Another interesting question is whether there exist extremal functions for the Trudinger-Moser inequality (1.1). The first breakthrough was due to the celebrated work of Carleson and Chang [4], who proved the existence of extremal functions when Ω is a ball in R n by the symmetrization argument. Later, Carleson and Chang's results were extended by Flucher [9] to arbitrary bounded domain in R 2 and by Lin [22] to general bounded domain in R n (n ≥ 2). Li [17,19] and [18] developed a blow-up method to establish the existence of extremal functions for the Trudinger-Moser inequality on Riemannian manifolds (see also [10,19,20,28,29,30]).
As we mentioned before, Wang and Ye [36] established Theorem B. In the spirit of Wang and Ye [36], in this paper, our main result is to establish an improved Hardy-Trudinger-Moser inequality and the existence of the extremals, which are stated as Theorem 1 and Theorem 2 respectively. The main tools we use are the rearrangement argument and blow-up analysis. Because of the presence of weights, we will encounter additional difficulties than the non-weighted case. Our main results are as follows.
Theorem 1.1. Let 0 < t < 2, u ∈ C ∞ 0 (B) and H(u) be defined as follows then there exists a constant C 0 > 0 such that the following inequality holds Here, we mention that the definition of H 1 (B) will be given in Section 2.
It's worthy to mention that when t = 0, Theorem 1 and Theorem 2 are the results of Wang and Ye [36].
Here, we describe the organization of this article. In Section 2, We reduce the problem to radially symmetric and nonincreasing function by the rearrangement argument. We prove the subcritical singular Hardy-Trudiger-Moser inequality and the existence of their extremals. In Section 3, we investigate the asymptotic behavior of the maximizing sequence {u k } near and far away from the origin by carrying out blow-up analysis procedure. Hence we complete the proof of Theorem 1.1. In Section 4, we assume the sequence {u k } blows up at the origin, then we give the explicit upper bound for critical function. In Section 5, by constructing proper test function sequence, we can verify that S exceeds the upper bound, which implies no blowing up occurs. Thus we complete the proof of Theorem 1.2.

Maximizers of subcritical version.
To prove Theorem 1, we first reduce the problem to radially symmetric decreasing functions by the rearrangement argument. For any u ∈ H 1 0 (B), where B is the unit disk in R 2 and H 1 0 (B) denotes the classical Sobolev space. Let u * denote the radially nonincreasing rearrangement with respect hence u H := H(u) defines a norm over H 1 0 (B). The completion of C ∞ 0 (B) with respect to the norm · H is a Hilbert space, which is denoted by H(B). For simplicity, we denote H(B) by H and · H by · . We know H(u) ≤ 1 implies H(u * ) ≤ 1.
We present the following subcritical version first. Lemma 2.1. Let 0 < t < 2, α k be an increasing sequence which converges to 4π(1 − t 2 ) as k → +∞, then there exists a constant C > 0 such that and the supremum can be attained by some u k ∈ H 1 (B).
Proof. We only need to show the inequality (2.1) holds for the subspace Σ. Moreover, it is enough to show the function Then as mentioned in the proof of Theorem 3 in [36], we claim that there exists two constants r 2 ∈ (0, 1) and We omit the detailed proof of this claim but refer the reader to [36]. For any r < r 2 , we have Here C k depends only on k. We use the singular Trudinger-Moser inequality on the bounded domain and get the following inequality Thus we complete the proof of inequality (2.1). Next, we will show the existence of extremals. Fix k, consider a maximizing sequence v j ∈ H 1 (B) for inequality ( It implies that e α k v 2 j |x| t converges to e α k u 2 k |x| t a.e. and is bounded in L q (B) for some q > 1. Naturally, we have Thus, u k is the extremal of sup Lemma 2.2. u k is a maximal sequence for S, that is to say Proof. On the one hand, it is obvious that lim 3. Proof of Theorem 1.1. We next consider the convergence of the sequence does not go to infinite as k → +∞, then there exists subsequence, which we still denote as {u k }, and u k ∞ ≤ C. We can assume u k u 0 in H and u k u 0 a.e. in B. Hence u 0 ≤ 1 and u 0 ∈ L ∞ (B). For any ω ∈ H and ω ≤ 1, the following inequality holds Exploiting monotone and dominated convergence theorem, we have That is to say In the following, we will suppose Theorem 1 does not hold true, then Next, we will use blow-up analysis to give a proof of Theorem 1 as in [17,18,19,28,29,30] and [36]. Since u k be the extremal function of the subcritical inequality, it is easy to check that u k = 1. By Euler-Lagrange multiplier theorem, one can easily calculate that {u k } satisfies the following Laplace equation Proof. Using the element inequality e t ≤ 1 + te t for t ≥ 0, one has By the proof of Lemma 2.2, we know that Proof. Suppose u 0 = 0, then there exists r 0 ∈ (0, 1 2 ) such that u 0 (r 0 ) > 0 and exists r 1 ∈ (0, r 0 ) and η > 0 such that when k is big enough, we have ∇u k L 2 (Br 1 ) ≤ 1 − η < 1. We omit the detailed statement here but refer the readers to Lemma 5 of [36]. By singular Trudinger-Moser inequality (1.2), we have Similarly to the proof of Lemma 5 in Wang and Ye [36], we can conclude that e 4π(1−t/2)u 2 k |·| t 1 ≤ C < +∞, which contradicts to (3.1). Thus u 0 ≡ 0.
Next, we define c k = u k (0) = max x∈B u k (x) and r 2 k = λ k c 2 k exp (α k c 2 k ) . By Lemma 2.1, we can obtain By the definition of r k and notice t < 2, we have Hence lim k→+∞ r k c k = 0 and lim Now we investigate the asymptotic behavior of {u k }. In view of 0 ≤ v k ≤ c k , for any Recall ξ(0) = 0 and ξ is radially symmetric and nonincreasing function, then the solution of equation (3.4) is uniquely determined as By the change of variables and careful calculation, we derive To investigate {u k } in the outer part and the rest region, we denote u k,L = min{u k , c k L }. We have the following lemma. Proof. One only need to notice the fact that  Proof. Firstly, we estimate e α k u 2 k |x| t 1 . Fix L > 2. On the one hand, note the fact that u k,L → 0 a.e. in B, we have On the other hand, by the uniform convergence of u k to zero in B c r for any r ∈ (0, 1), we obtain Consequently, lim k→+∞ I k = 2π 2−t when we take r → 0. Furthermore, In this way, we get the following which means lim inf k→+∞ λ −1 k c 2 k = 0. Hence we acquire equation (3.7). In order to prove the equation (3.8), we estimate the integral on three parts respectively. First, it is obvious that Then, for any R > 0, by the change of variables, we can get For the neck region, we have With the help of (3.6), we can complete the proof of equation (3.8) by tending R to +∞.

XUMIN WANG
Next, we need to investigate the convergence behavior of u k . Let g k = c k u k , then g k satisfies the equation (3.10) The equation (3.8) and its proof illustrates λ −1 k g k e α k u 2 k converges to the Dirac operator δ 0 in sense of measure, which suggests that g k should tend to the corresponding Green's function G 0 . We conformed this proposition as follows. Here we omit the detailed proof but refer the readers to Proposition 2 and Proposition 3 of the Wang and Ye's paper [36]. Proposition 1. The family g k converges to G 0 in W 1,p loc (B) weakly if p ∈ (1, 2), strongly in L q (B) for all q ≥ 1 and also in C(B c r ), ∀r ∈ (0, 1). Here G 0 is defined as follows Now, we give the proof of Theorem 1.
Proof. Proof of Theorem 1. Suppose Theorem 1 is false and let ρ ∈ (0, 1) be a small constant. By virtue of Proposition 1, we have As mentioned in Section 5 of [36], for a fixed ρ, it is easy to see that Bρ |∇u k | 2 dx < 1, for k big enough.
With the help of singular Trudinger-Moser inequality to (u k − u k (ρ)) + ∈ H 1 0 (B), we get On the other hand, there holds Obviously, this contradicts to the hypothesis (3.1). Hence we finish the proof of Theorem 1.

The upper bound of S.
Let {u k } be the maximum sequence of S. We will prove the existence of the extremal function by contradiction. Assume Theorem 1.2 is not valid, then lim k→+∞ u k ∞ = +∞. Define We already know S < +∞ by Theorem 1.1. And it is obvious S > 2π 2−t . All arguments and the properties acquired for u k in the previous context are true, except two properties. One is the proof of Lemma 3.3 and the other is Property (3.7).
We give a new approach of Lemma 3.3. Fix ρ ∈ (0, 1), in view of u k is uniformly bounded in B c ρ and lim k→+∞ c k = +∞, then for any L > 1, u k,L = u k in B c ρ for k big enough. It yields When L → +∞, we have u 0 = 0 in B c ρ . Notice ρ > 0 is arbitrary, consequently u 0 = 0.
We will show that Property (3.7) is not true. Actually, we have the next lemma.
Letting ρ → 1 and applying the uniform convergence of u k to 0 in B c r for r > 0, we complete the proof.
We proceed as in [36] to reach a contradiction. It is essential to give a claim of Carleson-Chang [4] type result.
where C G is given in (3.11).
Proof. Fix L > 2, we divide the integral areas into three regions. Firstly, it is easy to check Secondly, we have the following Thirdly, for the integral over the interior region B Rr 2 2−t k , we borrow some ideas from [36]. Fix a small constant ρ ∈ (0, 1), we have The first term goes to 0 when k → +∞, we get Then we know where lim k→+∞ o k (1) = 0 and Moreover, c −1 k l k → 1 uniformly in B Rr 2 2−t k since c −1 k u k → 1 uniformly in B R . As analyzed in [36], we have For any small ρ > 0, combing the three parts of estimation and let R go to +∞, we can get Employing the expansion of G 0 , we can get 5. The proof of Theorem 1.2. By constructing proper test function sequence, we can verify that S exceeds the upper bound, which implies no blowing up occurs. Then we complete the proof of Theorem 2 by contradiction.
Proof. We will construct a test function as follows where R = (− ln ε) 2 2−t , β ε and γ ε are constants to be chosen later. First, choose γ ε such that which satisfies the function continuous. We can get the following by expanding G 0 It is easy to see that f ε ∈ H. Now we estimate f ε . Let 0 < r < ρ < 1, as mentioned in [36], taking ρ → 1, we get On the other hand, by change of variable and integration by parts, we have